A particle travels along the X-axis as x = u(t - 2 s) + a(t - 2...

Given position $x = u(t - 2) + a(t - 2)^2$

Differentiating with respect to $t$, the velocity is $v = \dfrac{dx}{dt} = u + 2a(t - 2)$

Differentiating again, the acceleration is $\dfrac{dv}{dt} = 2a$

At $t = 0$, the initial velocity is $v = u + 2a(0 - 2) = u - 4a$

At $t = 2 \text{ s}$, the position is $x = u(2 - 2) + a(2 - 2)^2 = 0$, which means the particle is at the origin.

Thus, the acceleration is $2a$ and at $t = 2 \text{ s}$, the particle is at the origin.